cp's OEIS Frontend

A067998 a(n) = n^2 - 2*n.

%I A067998 #100 Jan 08 2025 05:48:11
%S A067998 0,-1,0,3,8,15,24,35,48,63,80,99,120,143,168,195,224,255,288,323,360,
%T A067998 399,440,483,528,575,624,675,728,783,840,899,960,1023,1088,1155,1224,
%U A067998 1295,1368,1443,1520,1599,1680,1763,1848,1935,2024,2115,2208,2303,2400,2499,2600,2703,2808,2915,3024,3135,3248,3363
%N A067998 a(n) = n^2 - 2*n.
%C A067998 a(n) is essentially the case 0 of the polygonal numbers. The polygonal numbers are defined as P_k(n) = Sum_{i=1..n} ((k-2)*i-(k-3)). Thus P_0(n) = 2*n - n^2 and a(n) = -P_0(n). - _Peter Luschny_, Jul 08 2011
%C A067998 For n >= 3, the denominator of the probability of winning the prize by switching from the initial choice of doors in a generalized Monty Hall problem with n doors: After a prize has been placed behind exactly one of the n doors at random, a contestant chooses a door. Then the host, who knows where the prize is, deliberately opens exactly one unchosen door that does not hide the prize (picked at random by the host among such doors when there is a choice) and then gives the contestant an opportunity to switch to any other door not yet opened. The numerator of this probability is n-1 (incidentally, gcd(n-1, n*(n-2)) = 1). The probability of winning by switching minus the probability of winning by not switching is thus (n-1)/(n*(n-2)) - 1/n = 1/a(n), which approaches zero as n approaches infinity, but nevertheless makes the switching strategy better for every finite n >= 3. The winning probability is 2/3 from switching in the classic 3-door Monty Hall problem; we have 3/8 and 4/15, respectively, in the 4- and 5-door generalizations. (The above analysis was independent but is consistent with the even more general "N-doors" section of the Wikipedia article, other parts of which make clear the historical importance of wording this problem as carefully as possible. See also A122774.) - _Rick L. Shepherd_, May 31 2014, clarified Oct 29 2015
%C A067998 For n > 1, a(n) is the largest integer k such that k + n^2 is a multiple of k + n. - _Derek Orr_, Sep 04 2014
%H A067998 Reinhard Zumkeller, <a href="/A067998/b067998.txt">Table of n, a(n) for n = 0..1000</a>
%H A067998 Wikipedia, <a href="http://www.wikipedia.org/wiki/Monty_Hall_problem">Monty Hall problem</a>.
%H A067998 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A067998 a(n) = A005563(n-2) = A005563(-n) = A000290(n-1)-1.
%F A067998 G.f.: x*(3*x-1)/(1-x)^3. - _Paul Barry_, Mar 27 2007
%F A067998 E.g.f.: exp(x)*(x^2-x). - _Paul Barry_, Mar 27 2007
%F A067998 a(n) = 2*n + a(n-1) - 3 (with a(0)=0). - _Vincenzo Librandi_, Aug 08 2010
%F A067998 From _Amiram Eldar_, Feb 17 2023: (Start)
%F A067998 Sum_{n>=3} 1/a(n) = 3/4.
%F A067998 Sum_{n>=3} (-1)^(n+1)/a(n) = 1/4. (End)
%p A067998 A067998:=n->n^2-2*n: seq(A067998(n), n=0..50); # _Wesley Ivan Hurt_, Sep 04 2014
%t A067998 Table[ n^2 - 2*n, {n, 0, 60} ] (* _George E. Antoniou_ *)
%t A067998 LinearRecurrence[{3, -3, 1}, {0, -1, 0}, 80] (* _Vladimir Joseph Stephan Orlovsky_, Feb 23 2012 *)
%o A067998 (PARI) a(n)=n^2-2*n;
%o A067998 (PARI) concat(0, Vec(x*(3*x-1)/(1-x)^3 + O(x^100))) \\ _Altug Alkan_, Oct 30 2015
%o A067998 (Haskell)
%o A067998 a067998 n = n * (n - 2)
%o A067998 a067998_list = scanl (+) 0 [-1, 1 ..]
%o A067998 -- _Reinhard Zumkeller_, Aug 26 2013
%o A067998 (Magma) [n^2-2*n : n in [0..50]]; // _Wesley Ivan Hurt_, Sep 04 2014
%Y A067998 Essentially the same as A005563.
%Y A067998 Cf. A060747 (first differences).
%Y A067998 Cf. A000290.
%K A067998 easy,sign
%O A067998 0,4
%A A067998 _George E. Antoniou_, Feb 06 2002
%E A067998 Edited and extended by _Robert G. Wilson v_, Feb 08 2002